What are the solutions to the equation (x^2-1)/4-(3x-1)/6 = x/2 ?

The solutions to the equation (x^2-1)/4-(3x-1)/6 = x/2 are x=(6+sqrt(39))/3 ,x=(6-sqrt(39))/3

Find the zeros of (x^2-1)/4-(3x-1)/6 = x/2

The zeros of (x^2-1)/4-(3x-1)/6 = x/2 are x=(6+sqrt(39))/3 ,x=(6-sqrt(39))/3

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\frac{x ^{2} - 1}{4} - \frac{3 x - 1}{6} = \frac{x}{2}

x = \frac{6 + 39}{3}, x = \frac{6 - 39}{3}

Decimal Notation

x = 4.08166 \dots, x = - 0.08166 \dots

Solution steps

\frac{x ^{2} - 1}{4} - \frac{3 x - 1}{6} = \frac{x}{2}

Find Least Common Multiplier of

4, 6, 2 : 12

Multiply by LCM=

12

\frac{x ^{2} - 1}{4} \cdot 12 - \frac{3 x - 1}{6} \cdot 12 = \frac{x}{2} \cdot 12

Simplify

3 (x^{2} - 1) - 2 (3 x - 1) = 6 x

Expand

3 (x^{2} - 1) - 2 (3 x - 1) : 3 x^{2} - 6 x - 1

3 x^{2} - 6 x - 1 = 6 x

Move

6 x

to the left side

3 x^{2} - 12 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 ( - 1 )}{2 \cdot 3}

(- 12)^{2} - 4 \cdot 3 (- 1) = 239

x_{1, 2} = \frac{- ( - 12 ) \pm 2 39}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 39}{2 \cdot 3}, x_{2} = \frac{- ( - 12 ) - 2 39}{2 \cdot 3}

x = \frac{- ( - 12 ) + 2 39}{2 \cdot 3} : \frac{6 + 39}{3}

x = \frac{- ( - 12 ) - 2 39}{2 \cdot 3} : \frac{6 - 39}{3}

The solutions to the quadratic equation are:

x = \frac{6 + 39}{3}, x = \frac{6 - 39}{3}

k^{2} + 3 k = 18

x^{2} - 2 = 14

x^{2} - 2 = 34

x^{2} = 1 7^{2} + 1 7^{2}

c, x = csc (y)

What are the solutions to the equation (x^2-1)/4-(3x-1)/6 = x/2 ?
The solutions to the equation (x^2-1)/4-(3x-1)/6 = x/2 are x=(6+sqrt(39))/3 ,x=(6-sqrt(39))/3
Find the zeros of (x^2-1)/4-(3x-1)/6 = x/2
The zeros of (x^2-1)/4-(3x-1)/6 = x/2 are x=(6+sqrt(39))/3 ,x=(6-sqrt(39))/3